Question

An oil company has two depots, A and B, with capacities of 7000 litres and 4000 litres respectively. The company is to supply oil to three petrol pumps, D, E, F whose requirements are 4500, 3000 and 3500 litres respectively. The distance (in km) between the depots and petrol pumps is given in the following table:

Figure

Assuming that the transportation cost per km is Rs 1.00 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?

Answer 1

Let x and y litres of oil be supplied from A to the petrol pumps D and E. Then, (7000 − x − y) L will be supplied from A to petrol pump F.

The requirement at petrol pump D is 4500 L. Since, x L are transported from depot A, the remaining (4500 −x) L will be transported from petrol pump B.

Similarly, (3000 − y) L and [3500 − (7000 − x − y)] L i.e. (x + y − 3500) L will be transported from depot B to petrol pump E and F. respectively.

The given problem can be represented diagrammatically as follows.

Since, quantity of oil are non-negative quantities.Therefore,

Cost of transporting 10 L of petrol = Re 1

Cost of transporting 1 L of petrol = $\mathrm{Rs}\frac{1}{10}$

Therefore, total transportation cost is given by,

The problem can be formulated as follows.

Minimize Z = 0.3x + 0.1y + 3950

subject to the constraints,

$$\begin{gathered} x+y\le 7000 \\ x\le 4500 \\ y\le 3000 \\ x+y\ge 3500 \\ x,y\ge 0 \end{gathered}$$

First we will convert inequations into equations as follows:
x + y = 7000, x = 4500, y = 3000, x + y = 3500, x = 0 and y = 0

Region represented by x + y ≤ 7000:
The line x + y = 7000 meets the coordinate axes at A₁(7000, 0) and $B_1\left(0,7000\right)$ respectively. By joining these points we obtain the line
x + y = 7000 . Clearly (0,0) satisfies the x + y = 7000 . So, the region which contains the origin represents the solution set of the inequation x + y ≤ 7000.

Region represented by x ​ ≤ 4500:
The line ​x = 4500 is the line passes through C₁(4500, 0) and is parallel to Y axis. The region to the left of the line x = 4500 will satisfy the inequation
x ​ ≤ 4500.

Region represented by y ​ ≤ 3000:
The line ​y = 3000 is the line passes through D₁(0, 3000) and is parallel to X axis. The region below the the line y = 3000 will satisfy the inequation
y ​ ≤ 3000.

Region represented by x + y ≥ 3500:
The line x + y = 7000 meets the coordinate axes at E₁(3500, 0) and $F_1\left(0,3500\right)$ respectively. By joining these points we obtain the line
x + y = 3500 . Clearly (0,0) satisfies the x + y = 3500. So, the region which contains the origin represents the solution set of the inequation x + y ≥ 3500.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints x + y ≤ 7000, x ​ ≤ 4500, y ​ ≤ 3000, x + y ≥ 3500, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the feasible region are E₁(3500, 0), C₁(4500, 0), I₁(4500, 2500), H₁(4000, 3000), and G₁(500, 3000).

The values of Z at these corner points are as follows.

Corner point	Z = 0.3x + 0.1y + 3950
E1(3500, 0)	5000
C1(4500, 0)	5300
I1(4500, 2500)	5550
H1(4000, 3000)	5450
G1(500, 3000)	4400

The minimum value of Z is 4400 at G₁(500, 3000).

Thus, the oil supplied from depot A is 500 L, 3000 L, and 3500 L and from depot B is 4000 L, 0 L, and 0 L to petrol pumps D, E, and F respectively.

The minimum transportation cost is Rs 4400.